Question

The equation of the circle passing through (1, 1) and the points of intersection of $x^2 + y^2 + 13x - 3y =0$ and $2x^2 + 2y^2 + 4 x - 7y -2 5 = 0$ is

• A. $4x^2 + 4y^2 - 30x - 10y = 25$
• B. $4x^2 + 4y^2 + 30x - 13y - 25 = 0$
• C. $4x^2 + 4y^2 - 17x - 10y + 25 = 0$
• D. None of the above

Answer 1

Answer: (B)

$4x^2 + 4y^2 + 30x - 13y - 25 = 0$

Answer 2

The required equation of circle is
$(x^2 + y^2 + 13x - 3y) + \lambda \Big( 11x+ \frac{1}{2} y+ \frac{25}{2} + ) =0 \, \, \, \, \, \, \, \, \, ....(i)$
Its passing through (1, 1)
$\Rightarrow \, \, \, \, \, \, \, \, \, \, 12 + \lambda(24) = 0$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \lambda = - \frac{1}{2}0$
On putting in E (i), we get
$x^2 + y^2 + 13x -3y - \frac{11}{2} x- \frac{1}{4}y-\frac{25}{4}=0$
$\Rightarrow \, \, \, \, 4X^2 + 4y^2 + 52 x - 12 y - 22 x - y - 25 =0$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, 4X^2 + 4y^2 + 30x -13y - 25 =0$